English

The function y = cx is the solution of differential equation dddydx=yx

Advertisements
Advertisements

Question

The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`

Options

  • True

  • False

MCQ
True or False
Advertisements

Solution

This statement is True.

shaalaa.com
  Is there an error in this question or solution?
Chapter 1.8: Differential Equation and Applications - Q.3

APPEARS IN

SCERT Maharashtra Mathematics and Statistics (Commerce) [English] 12 Standard HSC
Chapter 1.8 Differential Equation and Applications
Q.3 | Q 10

RELATED QUESTIONS

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.


\[\left( x^2 + 1 \right)\frac{dy}{dx} = 1\]

\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

\[\frac{dy}{dx} + \frac{1 + y^2}{y} = 0\]

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

xy (y + 1) dy = (x2 + 1) dx


\[y\sqrt{1 + x^2} + x\sqrt{1 + y^2}\frac{dy}{dx} = 0\]

\[\cos x \cos y\frac{dy}{dx} = - \sin x \sin y\]

\[\frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 , y\left( 0 \right) = 1\]

Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.


Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\]  given that y = 1, when x = 0.


\[\frac{dy}{dx} = \left( x + y + 1 \right)^2\]

\[\frac{dy}{dx}\cos\left( x - y \right) = 1\]

x2 dy + y (x + y) dx = 0


\[x^2 \frac{dy}{dx} = x^2 - 2 y^2 + xy\]

(y2 − 2xy) dx = (x2 − 2xy) dy


Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.


The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


Solve the following differential equation.

`dy /dx +(x-2 y)/ (2x- y)= 0`


Select and write the correct alternative from the given option for the question

The differential equation of y = Ae5x + Be–5x is


Choose the correct alternative:

General solution of `y - x ("d"y)/("d"x)` = 0 is


Solve the following differential equation `("d"y)/("d"x)` = x2y + y


Solve the following differential equation

`y log y ("d"x)/("d"y) + x` = log y


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


Solve: ydx – xdy = x2ydx.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×